Physics: Principles with Applications (7th Edition)

Published by Pearson
ISBN 10: 0-32162-592-7
ISBN 13: 978-0-32162-592-2

Chapter 26 - The Special Theory of Relativity - General Problems - Page 769: 57

Answer

$v=0.91c=2.7\times10^8m/s$

Work Step by Step

The proper time between heartbeats is $\Delta t_o=1.0 s$, or 60 beats per minute. Because of the astronaut’s motion, this interval is dilated to $\Delta t=2.4 s$ (i.e, 25 beats per minute). Solve for the speed from the time dilation relationship, equation 26–1a. $$\Delta t_o = \Delta t \sqrt{1-\frac{ v^{2} } { c^{2} }} $$ $$v=c\sqrt{1-(\frac{\Delta t_o}{\Delta t})^2}$$ $$v=c\sqrt{1-(\frac{1.0s}{2.4s })^2}$$ $$v=0.91c=2.7\times10^8m/s$$
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